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3 years ago

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Chemistry

1 answer:

Vilka [71]3 years ago

6 0

The period is the end of the sentence!!!

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An objects tendency to maintain its current state of motion is called

daser333 [38]

Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object's mass.

3 0

3 years ago

Which compound is classified as a hydrocarbon? A) butanal B) butyne C) 2-butanol D) 2-butanone

fenix001 [56]

Answer:

butyne

Explanation:

alkane, alkene, and alkyne are all examples of hydrocarbons.

butyne = alkyne

7 0

3 years ago

You have a 5-liter container with 1.30 x 1024 molecules of ammonia gas (NH3) at STP.

bogdanovich [222]

Answer:

3). 1.30 × 10^(24) molecules

Explanation:

From avogadro's law which state that equal volume of all gases at the same temperature and pressure contain the same number of molecules.

We can relate it to this question as;

V₁/n₁ = V₂/n₂

Where;

V₁ is initial volume

n₁ is initial number of molecules

V₂ is final volume

n₂ is final number of molecules

Thus at STP, we have V₁ = V₂ and as such Plugging in the relevant values gives;

5/(1.30 x 10^(24)) = 5/n₂

n₂ = 1.30 x 10^(24) molecules

6 0

2 years ago

What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)

Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

$Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\ ml\\\\V = 33.33 \ ml$

Therefore, volume required is 33.33 ml .

5 0

3 years ago

Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re

bearhunter [10]

Answer: 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles$

The balanced chemical equuation is:

$4Al+3O_2\rightarrow 2Al_2O_3$

According to stoichiometry :

4 moles of $Al$ produce == 2 moles of $Al_2O_3$

Thus 0.556 moles of $Al$ will produce= $\frac{2}{4}\times 0.556=0.278moles$ of $Al_2O_3$

Mass of $Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g$

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0

3 years ago